Kitonum

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These are replies submitted by Kitonum

@Carl Love  I know. Of course, if for OP is necessary to select the smallest elements, the sum of which does not exceed a bound, then we can use a set. Otherwise, we should use a list.

@Rouben Rostamian   If you take   large enough, (n-5)/6^6  will be greater than 1.

1. 4 is the the total number of the roots of the equation.

2. To solve this equation it is better to use  Student:-Calculus1:-Roots  command. In your example, it immediately returns all 4 roots in symbolic form:

Student:-Calculus1:-Roots((x-1)*(x^3-9*x^2+4), x);

@weidade37211  As a part of your equation, write  (1-10^(-15))^(10^10-1);  in a separate line and press enter (you get  overflow). This is a rational number with a giant numerator and denominator. Maple tries to calculate it exactly and overflow occurs. To avoid this, it is enough to put a decimal point on one of the exact numbers. Then all calculations will occur in float format with the accuracy established by Digits (with 10 digits by default).

(1-10^(-15))^(10^10-1);
Error, numeric exception: overflow

(1-10^(-15))^(10^10.-1);
                                                   1.

To avoid loss of accuracy, I increased  Digits  in my answer below. See:
Digits:=20:
(1-10^(-15))^(10^10.-1);
                                                 
0.99999000005000083332

@Wavish  Since you do not have a clear criterion for obtaining the wished line, then the solution "by eye" is likely to be the easiest and best solution to the problem.

An example of the creation of a specific matrix function:

F:=unapply(<t, t^2; sin(t), cos(t)>, t):
F(2);

@vv  In fact, as a formal symbolic result, we get the same thing. But if we try to assign some numeric constants to the symbols, then of course we get an error:

Ec := (Ems+I*Eml)*(1+((Ems+I*Eml)/Ef-1)*Zeta*phi/((Ems+I*Eml)/Ef+Zeta))/(1-((Ems+I*Eml)/Ef-1)*phi/((Ems+I*Eml)/Ef+Zeta)):
a:=simplify(Re(Ec)) assuming positive;
b:=simplify(Im(Ec)) assuming positive;
EC:=eval(Ec,Zeta=ZETA):
simplify(evalc([Re,Im](EC)));
eval(%, ZETA=Zeta);
is(a=%[1] and b=%[2]);
Zeta:=1;

 

@bsoudmand  This is the output of my code (in Maple 2018.2):


In the text form:

a:=(-Ems*Zeta*(Ef^2-2*Ef*Ems+Eml^2+Ems^2)*phi^2+((Zeta-1)*Ems^3+Ef*(Zeta-1)^2*Ems^2-(Zeta-1)*(Ef^2*Zeta-Eml^2)*Ems-Ef*Eml^2*(Zeta+1)^2)*phi+Ems*(Ef^2*Zeta^2+2*Ef*Ems*Zeta+Eml^2+Ems^2))/((Ef^2-2*Ef*Ems+Eml^2+Ems^2)*phi^2+(-2*Ems^2-2*Ef*(Zeta-1)*Ems+2*Ef^2*Zeta-2*Eml^2)*phi+Ef^2*Zeta^2+2*Ef*Ems*Zeta+Eml^2+Ems^2);

b:=-Eml*(Zeta*(Ef^2-2*Ef*Ems+Eml^2+Ems^2)*phi^2+(Ef*(Ef-2*Ems)*Zeta^2+(-Ef^2-Eml^2-Ems^2)*Zeta-2*Ef*Ems+Eml^2+Ems^2)*phi-Ef^2*Zeta^2-2*Ef*Ems*Zeta-Eml^2-Ems^2)/((Ef^2-2*Ef*Ems+Eml^2+Ems^2)*phi^2+(2*Ef*(-Ems+Ef)*Zeta+2*Ef*Ems-2*Eml^2-2*Ems^2)*phi+Ef^2*Zeta^2+2*Ef*Ems*Zeta+Eml^2+Ems^2);

@vv 

SymFun(ln((x-1)/(x+1)), x);
plot(ln((x-1)/(x+1)), x=-4..4, -4..4);

SymFun(sqrt(x^2), x);
plot(sqrt(x^2), x=-2..2);

SymFun(abs(x)^(1/2), x);
plot(abs(x)^(1/2), x=-1..1);

 

@emendes  You can use  makeproc  option. See help on  rsolve  command for this.

@Lali_miani  If you enter in 2d math mode any of  i, j, I  from the Comman Symbols palette, it will work as the imaginary unit. But from the keyboard you have to enter the imaginary unit as  I . I almost never use palettes and only work from the keyboard in 1d math mode (Maple input). 

@minhthien2016  You can make arrows at the ends of the axes of coordinates and labels in the same place if you use  plots:-arrow  and  plots:-textplot  commands.

Example:

Arrow_x:=plots:-arrow([2.3,0],[0.15,0], width=0, head_width=0.12, head_length=0.15):
Arrow_y:=plots:-arrow([0,4.3],[0,0.15], width=0, head_width=0.12, head_length=0.15):
Labels:=plots:-textplot([[2.4,-0.2,"x"],[-0.2,4.4,"y"]], font=[times,16]):
P:=plot(x^2, x=-2..2, color=red, thickness=2, labels=["",""]):
plots:-display(Arrow_x,Arrow_y,P,Labels);

Output:
                         

Edit.

@bliengme 

Add the line

convert(Q, exp);

@EB1000 The  DirectSearch  package is not included in Maple and must be downloaded from the Maple Application Center from  here

@tomleslie  Thanks for this. In my opinion it is the best solution.

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